If $x \diamond y = x-7y$ and $x \oplus y = 4x^{2}-y^{2}$, find $(3 \oplus 5) \diamond -4$.
Solution: First, find $3 \oplus 5$ $ 3 \oplus 5 = 4(3^{2})-5^{2}$ $ \hphantom{3 \oplus 5} = 11$ Now, find $11 \diamond -4$ $ 11 \diamond -4 = 11-(7)(-4)$ $ \hphantom{11 \diamond -4} = 39$.